Answer to Question #179931 in General Chemistry for Ally

Question #179931

A student is given 2 solutions: 15.0 ml of a 0.125 M KOH solution and 12.0 ml of a 0.175 M H2SO4 solution. Calculate the following:


a) The amount of potassium sulfate produced.


b) What is the limiting reactant? What is the excess reactant?


c) How many grams of the excess reactants are in excess?


1
Expert's answer
2021-04-11T23:54:16-0400

2KOH + H2SO4 → K2SO4 + 2H2O


Moles of KOH = 151000×0.125=1.875×103moles\frac{15}{1000}×0.125=1.875×10^{-3}moles


Amount of K2SO4=1.875×1032=9.375×104K_2SO_4=\frac{1.875×10^{-3}}{2}=9.375×10^{-4}


Limiting reactant is KOHKOH


Excess reactant is H2SO4H_2SO_4


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